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authorPrefetch2026-09-05 21:55:33 +0200
committerPrefetch2026-09-05 21:55:33 +0200
commit5cacf4ffaf3a9621ab536195f6469f98a420f054 (patch)
tree317b734468a287c50d2403fdfafa45434f538150 /source/know/concept/boltzmann-equation/index.md
parent29b49508a751649310173e592b63415dbf563a2a (diff)
Improve knowledge baseHEADmaster
Diffstat (limited to 'source/know/concept/boltzmann-equation/index.md')
-rw-r--r--source/know/concept/boltzmann-equation/index.md20
1 files changed, 12 insertions, 8 deletions
diff --git a/source/know/concept/boltzmann-equation/index.md b/source/know/concept/boltzmann-equation/index.md
index 5f4add0..3821512 100644
--- a/source/know/concept/boltzmann-equation/index.md
+++ b/source/know/concept/boltzmann-equation/index.md
@@ -67,7 +67,7 @@ but unfortunately also quite difficult to work with.
In addition, $$f$$ is a 7-dimensional function,
so the BTE is already hard to solve without collisions!
We only present the simplest case,
-known as the **Bhatnagar-Gross-Krook approximation**:
+the **Bhatnagar-Gross-Krook approximation**:
if the equilibrium state $$f_0(\vb{r}, \vb{v})$$ is known,
then each collision brings the system closer to $$f_0$$:
@@ -90,14 +90,15 @@ $$\begin{aligned}
n(\vb{r}, t) = \int_{-\infty}^\infty f(\vb{r}, \vb{v}, t) \dd{\vb{v}}
\end{aligned}$$
-Consequently, a purely velocity-dependent quantity $$Q(\vb{v})$$ can be averaged like so:
+Consequently, a purely velocity-dependent quantity $$Q(\vb{v})$$
+can be averaged like so:
$$\begin{aligned}
- \Expval{Q}
- = \frac{1}{n} \int_{-\infty}^\infty Q(\vb{r}, \vb{v}, t) \: f(\vb{r}, \vb{v}, t) \dd{\vb{v}}
+ \Expval{Q}\!(\vb{r}, t)
+ = \frac{1}{n} \int_{-\infty}^\infty Q(\vb{v}) \: f(\vb{r}, \vb{v}, t) \dd{\vb{v}}
\end{aligned}$$
-With that in mind, we multiply the collisionless BTE equation by $$Q(\vb{v})$$ and integrate,
+With that in mind, we multiply the collisionless BTE by $$Q(\vb{v})$$ and integrate,
assuming that $$\vb{F}$$ does not depend on $$\vb{v}$$:
$$\begin{aligned}
@@ -136,7 +137,8 @@ $$\begin{aligned}
If we set $$Q = m$$, then the mass density $$\rho = n \Expval{Q}$$,
and we find that the **zeroth moment** of the BTE describes conservation of mass,
-where $$\vb{V} \equiv \Expval{\vb{v}} = \int \vb{v} f \dd{\vb{v}}$$ is the fluid velocity:
+where $$\vb{V} \equiv \Expval{\vb{v}} = n^{-1} \int \vb{v} f \dd{\vb{v}}$$
+is the fluid velocity:
$$\begin{aligned}
\boxed{
@@ -231,7 +233,8 @@ $$\begin{aligned}
{% include proof/start.html id="proof-moment2" -%}
-We insert $$Q = m |\vb{v}|^2 / 2$$ into our prototype and recognize $$\rho$$ wherever possible:
+We insert $$Q = m |\vb{v}|^2 / 2$$ into our prototype
+and recognize $$\rho$$ wherever possible:
$$\begin{aligned}
0
@@ -244,7 +247,8 @@ $$\begin{aligned}
- \frac{\vb{F}}{2} \cdot \bigg( n \Expval{\pdv{|\vb{v}|^2}{\vb{v}}} \bigg)
\end{aligned}$$
-We handle these terms one by one. Substituting $$\vb{v} = \vb{V} + \vb{w}$$ in the first gives:
+We handle these terms one by one.
+Substituting $$\vb{v} = \vb{V} + \vb{w}$$ in the first gives:
$$\begin{aligned}
\Expval{|\vb{v}|^2}